What is the area of the region enclosed by the graphs of $f(x)=x^2+2x+11$, $g(x)=-4x+2$, and $x=0$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $63$ (Choice B) B $9$ (Choice C) C $6$ (Choice D) D $21$
Explanation: Visualizing the area We sketch the graphs of $f$ and $g$ first. ${\llap{-}1}$ ${\llap{-}2}$ ${\llap{-}3}$ ${2}$ ${4}$ ${6}$ ${8}$ ${10}$ ${12}$ ${14}$ $f$ $g$ $y$ $x$ From the graph, it appears that $f(x)\ge g(x)$ between the point where the graphs intersect and $x=0$. From this we are looking to evaluate: $ \int_{a}^{0}\left( f(x)-g(x) \right)\,dx$ where $a$ is the $x$ -coordinate of the point of intersection. Finding the $x$ -coordinate of the intersection point We can find the $x$ -coordinate of the point of intersection by setting the functions equal to each other and solving the resulting equation. $\begin{aligned} f(x)&=g(x) \\\\ x^2+2x+11 &=-4x+2\\\\ x^2+6x+9 &= 0 \\\\ (x+3)^2 &=0 \end{aligned}$ The graphs intersect where $x=-3$. Setting up the definite integral Thus, the area of the shaded region pictured above is given by: $\begin{aligned} &\phantom{=} \int_{-3}^{0}\left(x^2+2x+11 -(-4x+2)\right)\,dx \\\\ &= \int_{-3}^{0} \left( x^2+6x+9 \right) \,dx \end{aligned}$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{-3}^{0} \left( x^2+6x+9 \right) \,dx \\\\ &= \dfrac{x^3}{3} + 3x^2 + 9x~\Bigg|_{-3}^{0} \\\\ &= \left( 0+0+0 \right) - \left( -9+27-27 \right) \\\\ &=9 \end{aligned}$ Answer The area is $9$ square units.